Školjka

%V0 Let $x, y, z$ be real numbers each of whose absolute value is different from $\displaystyle \frac{1}{\sqrt 3}$ such that $x + y + z = xyz$. Prove that $$\frac{3x-x^{3}}{1-3x^{2}}+\frac{3y-y^{3}}{1-3y^{2}}+\frac{3z-z^{3}}{1-3z^{2}}=\frac{3x-x^{3}}{1-3x^{2}}\cdot\frac{3y-y^{3}}{1-3y^{2}}\cdot\frac{3z-z^{3}}{1-3z^{2}}$$